Electricity and magnetism were considered separate and unrelated phenomena for a long time. In the early decades of the nineteenth century, experiments on electric current by Oersted, Ampere and a few others established the fact that electricity and magnetism are inter-related. They found that moving electric charges produce magnetic fields. For example, an electric current deflects a magnetic compass needle placed in its vicinity. This naturally raises the questions like: Is the converse effect possible? Can moving magnets produce electric currents? Does the nature permit such a relation between electricity and magnetism? The answer is resounding yes! The experiments of Michael Faraday in England and Joseph Henry in USA, conducted around 1830, demonstrated conclusively that electric currents were induced in closed coils when subjected to changing magnetic fields. In this chapter, we will study the phenomena associated with changing magnetic fields and understand the underlying principles. The phenomenon in which electric current is generated by varying magnetic fields is appropriately called electromagnetic induction.

When Faraday first made public his discovery that relative motion between a bar magnet and a wire loop produced a small current in the latter, he was asked, “What is the use of it?” His reply was: “What is the use of a new born baby?” The phenomenon of electromagnetic induction is not merely of theoretical or academic interest but also of practical utility. Imagine a world where there is no electricity – no electric lights, no trains, no telephones and no personal computers. The pioneering experiments of Faraday and Henry have led directly to the development of modern day generators and transformers. Today’s civilisation owes its progress to a great extent to the discovery of electromagnetic induction.

**Josheph Henry [1797 – 1878]** American experimental physicist, professor at Princeton University and first director of the Smithsonian Institution. He made important improvements in electro- magnets by winding coils of insulated wire around iron pole pieces and invented an electromagnetic motor and a new, efficient telegraph. He discoverd self-induction and investigated how currents in one circuit induce currents in another.

The discovery and understanding of electromagnetic induction are based on a long series of experiments carried out by Faraday and Henry. We shall now describe some of these experiments.

**Experiment 6.1**

Figure 6.1 shows a coil C_{1}* connected to a galvanometer G. When the North-pole of a bar magnet is pushed towards the coil, the pointer in the galvanometer deflects, indicating the presence of electric current in the coil. The deflection lasts as long as the bar magnet is in motion. The galvanometer does not show any deflection when the magnet is held stationary. When the magnet is pulled away from the coil, the galvanometer shows deflection in the opposite direction, which indicates reversal of the current’s direction. Moreover, when the South-pole of the bar magnet is moved towards or away from the coil, the deflections in the galvanometer are opposite to that observed with the North-pole for similar movements. Further, the deflection (and hence current) is found to be larger when the magnet is pushed towards or pulled away from the coil faster. Instead, when the bar magnet is held fixed and the coil C_{1} is moved towards or away from the magnet, the same effects are observed. It shows that it is the relative motion between the magnet and the coil that is responsible for generation (induction) of electric current in the coil.

*Figure 6.1 When the bar magnet is pushed towards the coil, the pointer in the galvanometer G deflects.*

**Experiment 6.2**

In Fig. 6.2 the bar magnet is replaced by a second coil C_{2} connected to a battery. The steady current in the coil C_{2} produces a steady magnetic field. As coil C_{2} is moved towards the coil C_{1}, the galvanometer shows a deflection. This indicates that electric current is induced in coil C_{1}. When C_{2} is moved away, the galvanometer shows a deflection again, but this time in the opposite direction. The deflection lasts as long as coil C_{2} is in motion. When the coil C_{2} is held fixed and C_{1} is moved, the same effects are observed. Again, it is the relative motion between the coils that induces the electric current.

** Wherever the term ‘coil’ or ‘loop’ is used, it is assumed that they are made up of conducting material and are prepared using wires which are coated with insulating material.*

*Figure 6.2 Current is induced in coil C _{1} due to motion of the current carrying coil C_{2}*.

**Experiment 6.3**

The above two experiments involved relative motion between a magnet and a coil and between two coils, respectively. Through another experiment, Faraday showed that this relative motion is not an absolute requirement. Figure 6.3 shows two coils C_{1} and C_{2} held stationary. Coil C_{1} is connected to galvanometer G while the second coil C_{2} is connected to a battery through a tapping key K.

It is observed that the galvanometer shows a momentary deflection when the tapping key K is pressed. The pointer in the galvanometer returns to zero immediately. If the key is held pressed continuously, there is no deflection in the galvanometer. When the key is released, a momentory deflection is observed again, but in the opposite direction. It is also observed that the deflection increases dramatically when an iron rod is inserted into the coils along their axis.

*Figure 6.3 Experimental set-up for Experiment 6.3.*

Faraday’s great insight lay in discovering a simple mathematical relation to explain the series of experiments he carried out on electromagnetic induction. However, before we state and appreciate his laws, we must get familiar with the notion of magnetic flux, Φ B. Magnetic flux is defined in the same way as electric flux is defined in Chapter 1. Magnetic flux through a plane of area A placed in a uniform magnetic field B (Fig. 6.4) can be written as

Φ_{B} = **B**.**A** = BA cos θ (6.1)

where θ is angle between B and A. The notion of the area as a vector has been discussed earlier in Chapter 1. Equation (6.1) can be extended to curved surfaces and nonuniform fields.

If the magnetic field has different magnitudes and directions at various parts of a surface as shown in Fig. 6.5, then the magnetic flux through the surface is given by

${\mathrm{\varphi}}_{\mathrm{B}}={\mathrm{B}}_{1}.{\mathrm{dA}}_{1}+{\mathrm{B}}_{2}.{\mathrm{dA}}_{2}+...=\sum {\mathrm{B}}_{\mathrm{i}}.{\mathrm{dA}}_{\mathrm{i}}$ (6.2)

where ‘all’ stands for summation over all the area elements dA_{i} comprising the surface and B_{i} is the magnetic field at the area element dA_{i}. The SI unit of magnetic flux is weber (Wb) or tesla meter squared (Tm^{2}). Magnetic flux is a scalar quantity.

From the experimental observations, Faraday arrived at a conclusion that an emf is induced in a coil when magnetic flux through the coil changes with time. Experimental observations discussed in Section 6.2 can be explained using this concept.

The motion of a magnet towards or away from coil C_{1} in Experiment 6.1 and moving a current-carrying coil C_{2} towards or away from coil C_{1} in Experiment 6.2, change the magnetic flux associated with coil C_{1}. The change in magnetic flux induces emf in coil C_{1}. It was this induced emf which caused electric current to flow in coil C_{1} and through the galvanometer. A plausible explanation for the observations of Experiment 6.3 is as follows: When the tapping key K is pressed, the current in coil C_{2} (and the resulting magnetic field) rises from zero to a maximum value in a short time. Consequently, the magnetic flux through the neighbouring coil C_{1} also increases. It is the change in magnetic flux through coil C_{1} that produces an induced emf in coil C_{1}. When the key is held pressed, current in coil C_{2} is constant. Therefore, there is no change in the magnetic flux through coil C_{1} and the current in coil C_{1 }drops to zero. When the key is released, the current in C_{2} and the resulting magnetic field decreases from the maximum value to zero in a short time. This results in a decrease in magnetic flux through coil C_{1} and hence again induces an electric current in coil C_{1}*. The common point in all these observations is that the time rate of change of magnetic flux through a circuit induces emf in it. Faraday stated experimental observations in the form of a law called Faraday’s law of electromagnetic induction. The law is stated below.

** Note that sensitive electrical instruments in the vicinity of an electromagnet can be damaged due to the induced emfs (and the resulting currents) when the electromagnet is turned on or off.*

The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

Mathematically, the induced emf is given by

(6.3)

**Michael Faraday [1791–1867]** Faraday made numerous contributions to science, viz., the discovery of electromagnetic induction, the laws of electrolysis, benzene, and the fact that the plane of polarisation is rotated in an electric field. He is also credited with the invention of the electric motor, the electric generator and the transformer. He is widely regarded as the greatest experimental scientist of the nineteenth century.

The negative sign indicates the direction of ε and hence the direction of current in a closed loop. This will be discussed in detail in the next section.

In the case of a closely wound coil of N turns, change of flux associated with each turn, is the same. Therefore, the expression for the total induced emf is given by

(6.4)

The induced emf can be increased by increasing the number of turns N of a closed coil.

From Eqs. (6.1) and (6.2), we see that the flux can be varied by changing any one or more of the terms B, A and θ. In Experiments 6.1 and 6.2 in Section 6.2, the flux is changed by varying B. The flux can also be altered by changing the shape of a coil (that is, by shrinking it or stretching it) in a magnetic field, or rotating a coil in a magnetic field such that the angle θ between B and A changes. In these cases too, an emf is induced in the respective coils.

**Example 6.1** Consider Experiment 6.2. (a) What would you do to obtain a large deflection of the galvanometer? (b) How would you demonstrate the presence of an induced current in the absence of a galvanometer?

**Solution**

(a) To obtain a large deflection, one or more of the following steps can be taken: (i) Use a rod made of soft iron inside the coil C2, (ii) Connect the coil to a powerful battery, and (iii) Move the arrangement rapidly towards the test coil C1.

(b) Replace the galvanometer by a small bulb, the kind one finds in a small torch light. The relative motion between the two coils will cause the bulb to glow and thus demonstrate the presence of an induced current.

In experimental physics one must learn to innovate. Michael Faraday who is ranked as one of the best experimentalists ever, was legendary for his innovative skills.

**Example 6.2** A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time-interval.

**Solution** The angle θ made by the area vector of the coil with the magnetic field is 45°. From Eq. (6.1), the initial magnetic flux is

Φ = BA cos θ

Final flux, Φ_{min} = 0

The change in flux is brought about in 0.70 s. From Eq. (6.3), the magnitude of the induced emf is given by

And the magnitude of the current is

Note that the earth’s magnetic field also produces a flux through the loop. But it is a steady field (which does not change within the time span of the experiment) and hence does not induce any emf.

**Example 6.3**

A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth’s magnetic field at the place is 3.0 × 10^{–5} T.

**Solution**

Initial flux through the coil,

Φ_{B (initial) }= BA cos θ

= 3.0 × 10^{–5} × (π ×10^{–2}) × cos 0°

= 3π × 10^{–7}^{ }Wb

Final flux after the rotation,

Φ_{B (final)} = 3.0 × 10^{–5} × (π ×10^{–2}) × cos 180°

= –3π × 10^{–7}^{ } Wb

Therefore, estimated value of the induced emf is,

= 500 × (6π × 10^{–7}^{ })/0.25

= 3.8 × 10^{–3} V

I = ε/R = 1.9 × 10^{–3} A

Note that the magnitudes of ε and I are the estimated values. Their instantaneous values are different and depend upon the speed of rotation at the particular instant.

In 1834, German physicist Heinrich Friedrich Lenz (1804-1865) deduced a rule, known as Lenz’s law which gives the polarity of the induced emf in a clear and concise fashion. The statement of the law is:

*The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. *

The negative sign shown in Eq. (6.3) represents this effect. We can understand Lenz’s law by examining Experiment 6.1 in Section 6.2.1. In Fig. 6.1, we see that the North-pole of a bar magnet is being pushed towards the closed coil. As the North-pole of the bar magnet moves towards the coil, the magnetic flux through the coil increases. Hence current is induced in the coil in such a direction that it opposes the increase in flux. This is possible only if the current in the coil is in a counter-clockwise direction with respect to an observer situated on the side of the magnet. Note that magnetic moment associated with this current has North polarity towards the North-pole of the approaching magnet. Similarly, if the North-pole of the magnet is being withdrawn from the coil, the magnetic flux through the coil will decrease. To counter this decrease in magnetic flux, the induced current in the coil flows in clockwise direction and its South-pole faces the receding North-pole of the bar magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in flux.

What will happen if an open circuit is used in place of the closed loop in the above example? In this case too, an emf is induced across the open ends of the circuit. The direction of the induced emf can be found using Lenz’s law. Consider Figs. 6.6 (a) and (b). They provide an easier way to understand the direction of induced currents. Note that the direction shown by and indicate the directions of the induced currents.

*Figure 6.6 Illustration of Lenz’s law.*

A little reflection on this matter should convince us on the correctness of Lenz’s law. Suppose that the induced current was in the direction opposite to the one depicted in Fig. 6.6(a). In that case, the South-pole due to the induced current will face the approaching North-pole of the magnet. The bar magnet will then be attracted towards the coil at an ever increasing acceleration. A gentle push on the magnet will initiate the process and its velocity and kinetic energy will continuously increase without expending any energy. If this can happen, one could construct a perpetual-motion machine by a suitable arrangement. This violates the law of conservation of energy and hence can not happen.

Now consider the correct case shown in Fig. 6.6(a). In this situation, the bar magnet experiences a repulsive force due to the induced current. Therefore, a person has to do work in moving the magnet. Where does the energy spent by the person go? This energy is dissipated by Joule heating produced by the induced current.

**Example 6.4**

Figure 6.7 shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz’s law.

*Figure 6.7*

**Solution**

(i) The magnetic flux through the rectangular loop abcd increases, due to the motion of the loop into the region of magnetic field, The induced current must flow along the path bcdab so that it opposes the increasing flux.

(ii) Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacb, so as to oppose the change in flux.

(iii) As the magnetic flux decreases due to motion of the irregular shaped loop abcd out of the region of magnetic field, the induced current flows along cdabc, so as to oppose change in flux.

Note that there are no induced current as long as the loops are completely inside or outside the region of the magnetic field.

**Example 6.5**

(a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets?

(b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop

(i) when it is wholly inside the region between the capacitor plates (ii) when it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop.

(c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region (Fig. 6.8) to a field-free region with a constant velocity v. In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops.

*Figure 6.8*

(d) Predict the polarity of the capacitor in the situation described by Fig. 6.9.

*Figure 6.9*

(a) No. However strong the magnet may be, current can be induced only by changing the magnetic flux through the loop.

(b) No current is induced in either case. Current can not be induced by changing the electric flux.

(c) The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field region is not constant, hence induced emf will vary accordingly.

(d) The polarity of plate ‘A’ will be positive with respect to plate ‘B’ in the capacitor.

Let us consider a straight conductor moving in a uniform and time-independent magnetic field. Figure 6.10 shows a rectangular conductor PQRS in which the conductor PQ is free to move. The rod PQ is moved towards the left with a constant velocity v as shown in the figure. Assume that there is no loss of energy due to friction. PQRS forms a closed circuit enclosing an area that changes as PQ moves. It is placed in a uniform magnetic field B which is perpendicular to the plane of this system. If the length RQ = x and RS = l, the magnetic flux ΦB enclosed by the loop PQRS will be

ΦB = Blx

Since x is changing with time, the rate of change of flux Φ_{B }will induce an emf given by:

= (6.5)

where we have used dx/dt = –v which is the speed of the conductor PQ. The induced emf Blv is called motional emf. Thus, we are able to produce induced emf by moving a conductor instead of varying the magnetic field, that is, by changing the magnetic flux enclosed by the circuit.

It is also possible to explain the motional emf expression in Eq. (6.5) by invoking the Lorentz force acting on the free charge carriers of conductor PQ. Consider any arbitrary charge q in the conductor PQ. When the rod moves with speed v, the charge will also be moving with speed v in the magnetic field B. The Lorentz force on this charge is qvB in magnitude, and its direction is towards Q. All charges experience the same force, in magnitude and direction, irrespective of their position in the rod PQ.

The work done in moving the charge from P to Q is,

W = qvBl

Since emf is the work done per unit charge,

= Blv

This equation gives emf induced across the rod PQ and is identical to Eq. (6.5). We stress that our presentation is not wholly rigorous. But it does help us to understand the basis of Faraday’s law when the conductor is moving in a uniform and time-independent magnetic field.

On the other hand, it is not obvious how an emf is induced when a conductor is stationary and the magnetic field is changing – a fact which Faraday verified by numerous experiments. In the case of a stationary conductor, the force on its charges is given by

F = q (E + v × B) = qE (6.6)

since v = 0. Thus, any force on the charge must arise from the electric field term E alone. Therefore, to explain the existence of induced emf or induced current, we must assume that a time-varying magnetic field generates an electric field. However, we hasten to add that electric fields produced by static electric charges have properties different from those produced by time-varying magnetic fields. In Chapter 4, we learnt that charges in motion (current) can exert force/torque on a stationary magnet. Conversely, a bar magnet in motion (or more generally, a changing magnetic field) can exert a force on the stationary charge. This is the fundamental significance of the Faraday’s discovery. Electricity and magnetism are related.

**Example 6.6** A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring (Fig. 6.11). A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?

*Figure 6.11*

**Method I**

As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached. Using Eq. (6.5), the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by

. Hence,

Note that we have used v = ω r. This gives

ε

= 157 V

**Method II**

To calculate the emf, we can imagine a closed loop OPQ in which point O and P are connected with a resistor R and OQ is the rotating rod. The potential difference across the resistor is then equal to the induced emf and equals B × (rate of change of area of loop). If θ is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ is given by

where R is the radius of the circle. Hence, the induced emf is

ε ==

[Note:]

This expression is identical to the expression obtained by Method I and we get the same value of ε.

**Example 6.7**

A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field H_{E} at a place. If H_{E} = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel? Note that 1 G = 10^{–4} T.

Induced emf = (1/2) ω B R^{2}

= (1/2) × 4π × 0.4 × 10^{–4} × (0.5)^{2}

= 6.28 × 10^{–5} V

The number of spokes is immaterial because the emf’s across the spokes are in parallel.

In Section 6.5, we discussed qualitatively that Lenz’s law is consistent with the law of conservation of energy. Now we shall explore this aspect further with a concrete example.

Let r be the resistance of movable arm PQ of the rectangular conductor shown in Fig. 6.10. We assume that the remaining arms QR, RS and SP have negligible resistances compared to r. Thus, the overall resistance of the rectangular loop is r and this does not change as PQ is moved. The current I in the loop is,

= (6.7)

On account of the presence of the magnetic field, there will be a force on the arm PQ. This force I (l × B), is directed outwards in the direction opposite to the velocity of the rod. The magnitude of this force is,

F = I l B =

where we have used Eq. (6.7). Note that this force arises due to drift velocity of charges (responsible for current) along the rod and the consequent Lorentz force acting on them.

Alternatively, the arm PQ is being pushed with a constant speed v, the power required to do this is,

= (6.8)

The agent that does this work is mechanical. Where does this mechanical energy go? The answer is: it is dissipated as Joule heat, and is given by

$={\left(\frac{Blv}{r}\right)}^{2}r$

which is identical to Eq. (6.8).

Thus, mechanical energy which was needed to move the arm PQ is converted into electrical energy (the induced emf) and then to thermal energy.

There is an interesting relationship between the charge flow through the circuit and the change in the magnetic flux. From Faraday’s law, we have learnt that the magnitude of the induced emf is,

However,

Thus,

**Example 6.8** Refer to Fig. 6.12(a). The arm PQ of the rectangular conductor is moved from x = 0, outwards. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and is zero for x > b. Only the arm PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b, and is then moved back to x = 0 with constant speed v. Obtain expressions for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variation of these quantities with distance.

**(a)**

*Figure 6.12*

**Solution**

Let us first consider the forward motion from x = 0 to x = 2b

The flux Φ_{B} linked with the circuit SPQR is

The induced emf is,

When the induced emf is non-zero, the current I is (in magnitude)

**(b)**

*Figure 6.12*

The force required to keep the arm PQ in constant motion is I lB. Its direction is to the left. In magnitude

The Joule heating loss is

One obtains similar expressions for the inward motion from x = 2b to x = 0. One can appreciate the whole process by examining the sketch of various quantities displayed in Fig. 6.12(b).

So far we have studied the electric currents induced in well defined paths in conductors like circular loops. Even when bulk pieces of conductors are subjected to changing magnetic flux, induced currents are produced in them. However, their flow patterns resemble swirling eddies in water. This effect was discovered by physicist Foucault (1819-1868) and these currents are called eddy currents.

Consider the apparatus shown in Fig. 6.13. A copper plate is allowed to swing like a simple pendulum between the pole pieces of a strong magnet. It is found that the motion is damped and in a little while the plate comes to a halt in the magnetic field. We can explain this phenomenon on the basis of electromagnetic induction. Magnetic flux associated with the plate keeps on changing as the plate moves in and out of the region between magnetic poles. The flux change induces eddy currents in the plate. Directions of eddy currents are opposite when the plate swings into the region between the poles and when it swings out of the region.

*Figure 6.13 Eddy currents are generated in the copper plate, while entering and leaving the region of magnetic field.*

If rectangular slots are made in the copper plate as shown in Fig. 6.14, area available to the flow of eddy currents is less. Thus, the pendulum plate with holes or slots reduces electromagnetic damping and the plate swings more freely. Note that magnetic moments of the induced currents (which oppose the motion) depend upon the area enclosed by the currents (recall equation m = IA in Chapter 4).

This fact is helpful in reducing eddy currents in the metallic cores of transformers, electric motors and other such devices in which a coil is to be wound over metallic core. Eddy currents are undesirable since they heat up the core and dissipate electrical energy in the form of heat. Eddy currents are minimised by using laminations of metal to make a metal core. The laminations are separated by an insulating material like lacquer. The plane of the laminations must be arranged parallel to the magnetic field, so that they cut across the eddy current paths. This arrangement reduces the strength of the eddy currents. Since the dissipation of electrical energy into heat depends on the square of the strength of electric current, heat loss is substantially reduced.

*Figure 6.14 Cutting slots in the copper plate reduces the effect of eddy currents.*

Eddy currents are used to advantage in certain applications like:

(i) *Magnetic braking in trains**:* Strong electromagnets are situated above the rails in some electrically powered trains. When the electromagnets are activated, the eddy currents induced in the rails oppose the motion of the train. As there are no mechanical linkages, the braking effect is smooth.

(ii) *Electromagnetic damping*: Certain galvanometers have a fixed core made of nonmagnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly.

(iii)* Induction furnace*: Induction furnace can be used to produce high temperatures and can be utilised to prepare alloys, by melting the constituent metals. A high frequency alternating current is passed through a coil which surrounds the metals to be melted. The eddy currents generated in the metals produce high temperatures sufficient to melt it.

(iv) *Electric power meters*: The shiny metal disc in the electric power meter (analogue type) rotates due to the eddy currents. Electric currents are induced in the disc by magnetic fields produced by sinusoidally varying currents in a coil.

You can observe the rotating shiny disc in the power meter of your house.

Take two hollow thin cylindrical pipes of equal internal diameters made of aluminium and PVC, respectively. Fix them vertically with clamps on retort stands. Take a small cylinderical magnet having diameter slightly smaller than the inner diameter of the pipes and drop it through each pipe in such a way that the magnet does not touch the sides of the pipes during its fall. You will observe that the magnet dropped through the PVC pipe takes the same time to come out of the pipe as it would take when dropped through the same height without the pipe. Note the time it takes to come out of the pipe in each case. You will see that the magnet takes much longer time in the case of aluminium pipe. Why is it so? It is due to the eddy currents that are generated in the aluminium pipe which oppose the change in magnetic flux, i.e., the motion of the magnet. The retarding force due to the eddy currents inhibits the motion of the magnet. Such phenomena are referred to as electromagnetic damping. Note that eddy currents are not generated in PVC pipe as its material is an insulator whereas aluminium is a conductor.

An electric current can be induced in a coil by flux change produced by another coil in its vicinity or flux change produced by the same coil. These two situations are described separately in the next two sub-sections. However, in both the cases, the flux through a coil is proportional to the current. That is, Φ_{B} α I.

Further, if the geometry of the coil does not vary with time then,

For a closely wound coil of N turns, the same magnetic flux is linked with all the turns. When the flux Φ_{B} through the coil changes, each turn contributes to the induced emf. Therefore, a term called flux linkage is used which is equal to NΦ_{B} for a closely wound coil and in such a case

$N{\Phi}_{B}\propto I$

The constant of proportionality, in this relation, is called inductance. We shall see that inductance depends only on the geometry of the coil and intrinsic material properties. This aspect is akin to capacitance which for a parallel plate capacitor depends on the plate area and plate separation (geometry) and the dielectric constant K of the intervening medium (intrinsic material property).

Inductance is a scalar quantity. It has the dimensions of [M L^{2 }T^{–2 }A^{–2}] given by the dimensions of flux divided by the dimensions of current. The SI unit of inductance is henry and is denoted by H. It is named in honour of Joseph Henry who discovered electromagnetic induction in USA, independently of Faraday in England.

In the previous sub-section, we considered the flux in one solenoid due to the current in the other. It is also possible that emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil. This phenomenon is called self-induction. In this case, flux linkage through a coil of N turns is proportional to the current through the coil and is expressed as

(6.15)

where constant of proportionality L is called self-inductance of the coil. It is also called the coefficient of self-induction of the coil. When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. Using Eq. (6.15), the induced emf is given by

(6.16)

Thus, the self-induced emf always opposes any change (increase or decrease) of current in the coil.

It is possible to calculate the self-inductance for circuits with simple geometries. Let us calculate the self-inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length. The magnetic field due to a current I flowing in the solenoid is B = µ_{0}n I (neglecting edge effects, as before). The total flux linked with the solenoid is

where nl is the total number of turns. Thus, the self-inductance is,

(6.17)

If we fill the inside of the solenoid with a material of relative permeability µ_{r} (for example soft iron, which has a high value of relative permeability), then,

(6.18)

The self-inductance of the coil depends on its geometry and on the permeability of the medium.

The self-induced emf is also called the back emf as it opposes any change in the current in a circuit. Physically, the self-inductance plays the role of inertia. It is the electromagnetic analogue of mass in mechanics. So, work needs to be done against the back emf (ε) in establishing the current. This work done is stored as magnetic potential energy. For the current I at an instant in a circuit, the rate of work done is

If we ignore the resistive losses and consider only inductive effect, then using Eq. (6.16),

Total amount of work done in establishing the current I is

$W=\int dW={\int}_{0}^{I}LIdI$

Thus, the energy required to build up the current I is,

(6.19)

This expression reminds us of mv^{2}/2 for the (mechanical) kinetic energy of a particle of mass m, and shows that L is analogous to m (i.e., L is electrical inertia and opposes growth and decay of current in the circuit).

Consider the general case of currents flowing simultaneously in two nearby coils. The flux linked with one coil will be the sum of two fluxes which exist independently. Equation (6.9) would be modified into

${N}_{1}{\varphi}_{1}={M}_{11}{I}_{1}+{M}_{12}{I}_{2}$

where M_{11} represents inductance due to the same coil.

Therefore, using Faraday’s law,

M_{11} is the self-inductance and is written as L_{1}. Therefore,

**Example 6.10** (a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid. (b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor?

**Solution**

(a) From Eq. (6.19), the magnetic energy is

$=\frac{1}{2}L{\left(\frac{{B}^{2}}{{\mu}_{o}n}\right)}^{2}(\mathrm{sin}ceB={\mu}_{o}nIforasolenoid)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left({\mu}_{o}{n}^{2}Al\right){\left(\frac{{B}^{2}}{{\mu}_{o}n}\right)}^{2}[fromEq.(6.17\left)\right]$

(b) The magnetic energy per unit volume is,

(where V is volume that contains flux)

(6.20)

We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor (refer to Chapter 2, Eq. 2.77),

(2.77)

In both the cases energy is proportional to the square of the field strength. Equations (6.20) and (2.77) have been derived for special cases: a solenoid and a parallel plate capacitor, respectively. But they are general and valid for any region of space in which a magnetic field or/and an electric field exist.

Consider Fig. 6.15 which shows two long co-axial solenoids each of length l. We denote the radius of the inner solenoid S_{1} by r_{1} and the number of turns per unit length by n_{1}. The corresponding quantities for the outer solenoid S_{2 }are r_{2 }and n_{2}, respectively. Let N_{1} and N_{2} be the total number of turns of coils S_{1} and S_{2}, respectively.

When a current I_{2} is set up through S_{2}, it in turn sets up a magnetic flux through S_{1}. Let us denote it by Φ_{1}. The corresponding flux linkage with solenoid S_{1} is

${N}_{1}{\varphi}_{1}={M}_{12}{I}_{2}(6.9)$

M_{12} is called the mutual inductance of solenoid S_{1} with respect to solenoid S_{2}. It is also referred to as the coefficient of mutual induction.

For these simple co-axial solenoids it is possible to calculate M_{12} . The magnetic field due to the current I_{2 }in S_{2} is µ_{0}n_{2}I_{2}. The resulting flux linkage with coil S_{1} is,

(6.10) where n_{1}l is the total number of turns in solenoid S_{1}. Thus, from Eq. (6.9) and Eq. (6.10),

M_{12} = µ_{0}n_{1}n_{2}πr_{1}^{2 }l (6.11)

Note that we neglected the edge effects and considered the magnetic field µ_{0}n_{2}I_{2} to be uniform throughout the length and width of the solenoid S_{2}. This is a good approximation keeping in mind that the solenoid is long, implying l >> r_{2}.

We now consider the reverse case. A current I_{1 }is passed through the solenoid S_{1} and the flux linkage with coil S_{2} is,

N_{2}Φ_{2} = M_{21} I_{1} (6.12)

M_{21} is called the mutual inductance of solenoid S_{2} with respect to solenoid S_{1}.

The flux due to the current I_{1} in S_{1} can be assumed to be confined solely inside S_{1} since the solenoids are very long. Thus, flux linkage with solenoid S_{2} is

*Figure 6.15 Two long co-axial solenoids of same length l.*

where n_{2}l is the total number of turns of S_{2}. From Eq. (6.12),

M_{21} = µ_{0}n_{1}n_{2}πr_{1}^{2}l (6.13)

Using Eq. (6.11) and Eq. (6.12), we get

M_{12 }= M_{21}= M (say) (6.14)

We have demonstrated this equality for long co-axial solenoids. However, the relation is far more general. Note that if the inner solenoid was much shorter than (and placed well inside) the outer solenoid, then we could still have calculated the flux linkage N_{1}Φ_{1} because the inner solenoid is effectively immersed in a uniform magnetic field due to the outer solenoid. In this case, the calculation of M_{12} would be easy. However, it would be extremely difficult to calculate the flux linkage with the outer solenoid as the magnetic field due to the inner solenoid would vary across the length as well as cross section of the outer solenoid. Therefore, the calculation of M_{21} would also be extremely difficult in this case. The equality M_{12}=M_{21} is very useful in such situations.

We explained the above example with air as the medium within the solenoids. Instead, if a medium of relative permeability µ_{r} had been present, the mutual inductance would be

M =µ_{r} µ_{0} n_{1}n_{2}π r^{2}_{1} l

It is also important to know that the mutual inductance of a pair of coils, solenoids, etc., depends on their separation as well as their relative orientation.

**Example**** 6.9** Two concentric circular coils, one of small radius r_{1} and the other of large radius r_{2}, such that r_{1} << r_{2}, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement.

**Solution** Let a current I_{2} flow through the outer circular coil. The field at the centre of the coil is B_{2} = µ_{0}I_{2} / 2r_{2}. Since the other

co-axially placed coil has a very small radius, B_{2} may be considered constant over its cross-sectional area. Hence,

Φ1 = πr_{1}^{2}B_{2}

= M_{12} I_{2}

Thus,

From Eq. (6.14)

Note that we calculated M_{12} from an approximate value of Φ_{1}, assuming the magnetic field B_{2} to be uniform over the area πr_{1}^{2}. However, we can accept this value because r_{1} << r_{2}.

Now, let us recollect Experiment 6.3 in Section 6.2. In that experiment, emf is induced in coil C_{1} wherever there was any change in current through coil C_{2}. Let Φ_{1} be the flux through coil C_{1} (say of N_{1} turns) when current in coil C_{2} is I_{2}.

Then, from Eq. (6.9), we have

N_{1}Φ_{1} = MI_{2 }

For currents varrying with time,

Since induced emf in coil C_{1} is given by

We get,

It shows that varying current in a coil can induce emf in a neighbouring coil. The magnitude of the induced emf depends upon the rate of change of current and mutual inductance of the two coils.

The phenomenon of electromagnetic induction has been technologically exploited in many ways. An exceptionally important application is the generation of alternating currents (ac). The modern ac generator with a typical output capacity of 100 MW is a highly evolved machine. In this section, we shall describe the basic principles behind this machine. The Yugoslav inventor Nicola Tesla is credited with the development of the machine. As was pointed out in Section 6.3, one method to induce an emf or current in a loop is through a change in the loop’s orientation or a change in its effective area. As the coil rotates in a magnetic field B, the effective area of the loop (the face perpendicular to the field) is A cos θ, where θ is the angle between A and B. This method of producing a flux change is the principle of operation of a simple ac generator. An ac generator converts mechanical energy into electrical energy.

The basic elements of an ac generator are shown in Fig. 6.16. It consists of a coil mounted on a rotor shaft. The axis of rotation of the coil is perpendicular to the direction of the magnetic field. The coil (called armature) is mechanically rotated in the uniform magnetic field by some external means. The rotation of the coil causes the magnetic flux through it to change, so an emf is induced in the coil. The ends of the coil are connected to an external circuit by means of slip rings and brushes.

When the coil is rotated with a constant angular speed ω, the angle θ between the magnetic field vector B and the area vector A of the coil at any instant t is θ = ωt (assuming θ = 0° at t = 0). As a result, the effective area of the coil exposed to the magnetic field lines changes with time, and from Eq. (6.1), the flux at any time t is

Φ_{B} = BA cos θ = BA cos ωt

From Faraday’s law, the induced emf for the rotating coil of N turns is then,

*Figure 6.16 AC Generator*

Thus, the instantaneous value of the emf is

(6.21)

where NBAω is the maximum value of the emf, which occurs when sin ωt = ±1. If we denote NBAω as ε_{0}, then

ε = ε_{0} sin ωt (6.22)

Since the value of the sine fuction varies between +1 and –1, the sign, or polarity of the emf changes with time. Note from Fig. 6.17 that the emf has its extremum value when θ = 90° or θ = 270°, as the change of flux is greatest at these points.

The direction of the current changes periodically and therefore the current is called alternating current (ac). Since ω = 2πν, Eq (6.22) can be written as

ε = ε_{0}sin 2π ν t (6.23)

where ν is the frequency of revolution of the generator’s coil.

Note that Eq. (6.22) and (6.23) give the instantaneous value of the emf and ε varies between +ε_{0} and –ε_{0} periodically. We shall learn how to determine the time-averaged value for the alternating voltage and current in the next chapter.

*Figure 6.17 An alternating emf is generated by a loop of wire rotating in a magnetic field.*

In commercial generators, the mechanical energy required for rotation of the armature is provided by water falling from a height, for example, from dams. These are called hydro-electric generators. Alternatively, water is heated to produce steam using coal or other sources. The steam at high pressure produces the rotation of the armature. These are called thermal generators. Instead of coal, if a nuclear fuel is used, we get nuclear power generators. Modern day generators produce electric power as high as 500 MW, i.e., one can light up 5 million 100 W bulbs! In most generators, the coils are held stationary and it is the electromagnets which are rotated. The frequency of rotation is 50 Hz in India. In certain countries such as USA, it is 60 Hz.

**Example 6.11** Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 turn coil of area 0.10 m^{2}. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil?

**Solution **Here ν = 0.5 Hz; N =100, A = 0.1 m2 and B = 0.01 T. Employing Eq. (6.21)

ε_{0} = NBA (2 π ν)

= 100 × 0.01 × 0.1 × 2 × 3.14 × 0.5

= 0.314 V

The maximum voltage is 0.314 V.

We urge you to explore such alternative possibilities for power generation.

The migratory pattern of birds is one of the mysteries in the field of biology, and indeed all of science. For example, every winter birds from Siberia fly unerringly to water spots in the Indian subcontinent. There has been a suggestion that electromagnetic induction may provide a clue to these migratory patterns. The earth’s magnetic field has existed throughout evolutionary history. It would be of great benefit to migratory birds to use this field to determine the direction. As far as we know birds contain no ferromagnetic material. So electromagnetic induction seems to be the only reasonable mechanism to determine direction. Consider the optimal case where the magnetic field B, the velocity of the bird v, and two relevant points of its anatomy separated by a distance l, all three are mutually perpendicular. From the formula for motional emf, Eq. (6.5),

ε = Blv

Taking B = 4 × 10–5 T, l = 2 cm wide, and v = 10 m/s, we obtain

ε = 4 × 10–5 × 2 × 10–2 × 10 V = 8 × 10–6 V

= 8 µV

This extremely small potential difference suggests that our hypothesis is of doubtful validity. Certain kinds of fish are able to detect small potential differences. However, in these fish, special cells have been identified which detect small voltage differences. In birds no such cells have been identified. Thus, the migration patterns of birds continues to remain a mystery.

1. The magnetic flux through a surface of area A placed in a uniform magnetic field B is defined as,

ΦB = B.A = BA cos θ

where θ is the angle between B and A.

2. Faraday’s laws of induction imply that the emf induced in a coil of N turns is directly related to the rate of change of flux through it,

Here ΦΒ is the flux linked with one turn of the coil. If the circuit is closed, a current I = ε/R is set up in it, where R is the resistance of the circuit.

3. Lenz’s law states that the polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces it. The negative sign in the expression for Faraday’s law indicates this fact.

4. When a metal rod of length l is placed normal to a uniform magnetic field B and moved with a velocity v perpendicular to the field, the induced emf (called motional emf) across its ends is

ε = Bl v

5. Changing magnetic fields can set up current loops in nearby metal (any conductor) bodies. They dissipate electrical energy as heat. Such currents are eddy currents.

6. Inductance is the ratio of the flux-linkage to current. It is equal to NΦ/I.

7. A changing current in a coil (coil 2) can induce an emf in a nearby coil (coil 1). This relation is given by,

The quantity M12 is called mutual inductance of coil 1 with respect to coil 2. One can similarly define M21. There exists a general equality,

M12 = M21

8. When a current in a coil changes, it induces a back emf in the same coil. The self-induced emf is given by,

L is the self-inductance of the coil. It is a measure of the inertia of the coil against the change of current through it.

9. The self-inductance of a long solenoid, the core of which consists of a magnetic material of permeability µr, is given by

L = µr µ0 n2 Al

where A is the area of cross-section of the solenoid, l its length and n the number of turns per unit length.

10. In an ac generator, mechanical energy is converted to electrical energy by virtue of electromagnetic induction. If coil of N turn and area A is rotated at ν revolutions per second in a uniform magnetic field B, then the motional emf produced is

ε = NBA (2πν) sin (2πνt)

where we have assumed that at time t = 0 s, the coil is perpendicular to the field.

**Points to Ponder**

1. Electricity and magnetism are intimately related. In the early part of the nineteenth century, the experiments of Oersted, Ampere and others established that moving charges (currents) produce a magnetic field. Somewhat later, around 1830, the experiments of Faraday and Henry demonstrated that a moving magnet can induce electric current.

2. In a closed circuit, electric currents are induced so as to oppose the changing magnetic flux. It is as per the law of conservation of energy. However, in case of an open circuit, an emf is induced across its ends. How is it related to the flux change?

3. The motional emf discussed in Section 6.5 can be argued independently from Faraday’s law using the Lorentz force on moving charges. However, even if the charges are stationary [and the q (v × B) term of the Lorentz force is not operative], an emf is nevertheless induced in the presence of a time-varying magnetic field. Thus, moving charges in static field and static charges in a time-varying field seem to be symmetric situation for Faraday’s law. This gives a tantalising hint on the relevance of the principle of relativity for Faraday’s law.

4. The motion of a copper plate is damped when it is allowed to oscillate between the magnetic pole-pieces. How is the damping force, produced by the eddy currents?

Exercise

**Q 1. Predict the direction of induced current in the Situations described by the Figures. 6.18(a) to (f).**

Question 6.2:

Use Lenz's law to determine the direction of induced current in the situations described by Fig.

(a) A wire of irregular shape turning into a circular shape

(b) A circular loop being deformed into a narrow straight wire.

**Q 3. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm ^{2} placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?**

**Q 4. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?**

**Q 5. We are rotating a 1 m long metallic rod with an angular frequency of 400 red $s-1$ with an axis normal to the rod passing through its one end. And on to the other end of the rod, it is connected with a circular metallic ring. There exist a uniform magnetic field of 0.5 T which is parallel to the axis everywhere. Find out the emf induced between the center and the ring.**

**Q 6. A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s–1 in a uniform horizontal magnetic field of magnitude 3.0 × 10–2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?**

**Q 7. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s ^{–1}, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10^{–4} Wb m^{–2}.**

(a) What is the instantaneous value of the emf induced in the wire?

(b) What is the direction of the emf?

(c) Which end of the wire is at the higher electrical potential?

**Q 8. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.**

**Q 9. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?**

**Q 10. A jet plane is traveling towards the west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10 ^{–4} T and the dip angle is 30°.**

**Additional Questions**

**Q 11. Let us assume that the loop in question number 4 is stationary or constant but the current source which is feeding the electromagnet which is producing the magnetic field is slowly decreased. It was having an initial value of 0.3 T and the rate of reducing the field is 0.02 T / sec. If the cut is joined to form the loop having a resistance of $1.6Ω$. Calculate how much power is lost in the form of heat? What is the source of this power?**

** Q 12. We have a square loop having side as 12 cm and its sides are parallel to x and the y-axis is moved with a velocity of 8 cm/s in the positive x-direction in a region which have a magnetic field in the direction of positive z-axis. The field is not uniform whether in case of its space or in the case of time. It has a gradient of 10 ^{-3 }Tcm^{-1} along the negative x-direction(i.e its value increases by 10^{-3 }Tcm^{-1} as we move from positive to negative direction ), and it is reducing in the case of time with the rate of 10^{-3 }Tcm^{-1}**

**Q 13. We have a powerful loudspeaker magnet and have to measure the magnitude of the field between the poles of the speaker. And a small search coil is placed normal to the field direction and then quickly removed out of the field region, the coil is of the area and has 25 closely wound turns. Similarly, we can give the coil a quick $9_{∘}$ turn to bring its plane parallel to the field direction. We have measured the total charge flown in the coil by using a ballistic galvanometer and it comes to 7.5 mC. Total resistance after combining the coil and the galvanometer is $0.50Ω$. Estimate the field strength of the magnet.**

**Q 14. In the given figure we have a metal rod PQ which is put on the smooth rails AB and these are kept in between the two poles of permanent magnets. All these three (rod, rails, and the magnetic field ) are in mutually perpendicular direction. There is a galvanometer ‘G’ connected through the rails by using a switch ‘K’.Given, Rod’s length = 15 cm , Magnetic field strength, B = 0.50 T, Resistance produced by the closed-loop = $9.0mΩ$. Let’s consider the field is uniform.**

**(i) Determine the polarity and the magnitude of the induced emf if we will keep the K open and the rod will be moved with the speed of 12 cm/s in the direction shown in the figure.**

** ****(ii) When the K was open is there any excess charge built up? Assume that K is closed then what will happen after it?**

** ****(iii) When the rod was moving uniformly and the K was open, then on the electron in the rod PQ there was no net force even though they did not experience any magnetic field because of the motion of the rod. Explain.**

** ****(iv) After closing the K, calculate the retarding force.**

** ****(v) When the K will be closed calculate the total external power which will be required to keep moving the rod with the same speed ( 12 cm/s)? and also calculate the power required when K will be closed.**

** ****(vi)What would be the power loss ( in form of heat) when the circuit is closed? What would be the source of this power?**

** ****(vii) Calculate the emf induced in the moving rod if the direction of the magnetic field is changed from perpendicular to parallel to the rails?**

**Q 15. We have an air-cored solenoid having a length of 30 cm, whose area is 25 cm ^{2}, and the number of turns is 500. And the solenoid has carried a current of 2.5 A. Suddenly the current is turned off and the time is taken for it is 10^{–3} s. What would be the average value of the induced back-emf by the ends of the open switch in the circuit? (Neglect the variation in the magnetic fields near the ends of the solenoid.)**

**Q 16.** **(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.**

**(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v= 10 m/s.Calculate the induced emf in the loop at the instant when x = 0.2 m.Take a = 0.1 m and assume that the loop has a large resistance.**

**Q 17.A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,**

B=−B_{0}k(r≤a;a<R)

**= 0 (otherwise)**

**What is the angular velocity of the wheel after the field is suddenly switched off?**

Q. 1. A square of side L meters lies in the xy-plane in a region, where the magnetic field is given by B=${\mathrm{B}}_{0}\left(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right)\mathrm{T},\mathrm{where}{\mathrm{B}}_{0}$ is constant. The magnetic of flux passing through the square is

$\left(\mathrm{a}\right)2{\mathrm{B}}_{0}{\mathrm{L}}^{2}\mathrm{Wb}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)3{\mathrm{B}}_{0}{\mathrm{L}}^{2}\mathrm{Wb}\phantom{\rule{0ex}{0ex}}\left(\mathrm{c}\right)4{\mathrm{B}}_{0}{\mathrm{L}}^{2}\mathrm{Wb}\phantom{\rule{0ex}{0ex}}\left(\mathrm{d}\right)\sqrt{29}{\mathrm{B}}_{0}{\mathrm{L}}^{2}\mathrm{Wb}$

NEETprep AnswerQ. 2. A loop, made of straight edges has six corners at A(0, 0, 0), B(L, 0, 0), C(L, L, 0), D(0, L, 0), E(0, L, L) and F(0, 0, L). A magnetic field B=${\mathrm{B}}_{0}\left(\hat{\mathrm{i}}+\hat{\mathrm{k}}\right)$ T is present in the region. The flux passing through the loop ABCDEFA (in that order) is

$\left(a\right){B}_{0}{L}^{2}Wb\phantom{\rule{0ex}{0ex}}\left(b\right)2{B}_{0}{L}^{2}Wb\phantom{\rule{0ex}{0ex}}\left(c\right)\sqrt{2}{B}_{0}{L}^{2}Wb\phantom{\rule{0ex}{0ex}}\left(d\right)4{B}_{0}{L}^{2}Wb$

NEETprep Answer

Q. 3. A cylindrical bar magnet is rotated about its axis. A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then,

1. a direct current flows in the ammeter A

2. no current flows through the ammeter A

3. an alternating sinusoidal current flows through the ammeter A with a time period$\mathrm{T}=\frac{2\mathrm{\pi}}{\mathrm{\omega}}$

4. a time varying non-sinusoidal current flows through the ammeter A

Q. 4. There are two coils A and B as shown in figure. A current starts flowing in B as shown, when A is moved towards B and stops when A stops moving. The current in A is counter clockwise. B is kept stationary when A moves. We can infer that

1. there is a constant current in the clockwise direction in A

2. there is a varying current in A

3. there is no current in A

4. there is a constant current in the counter clockwise direction in A

Q. 5. Same as problem 4 except the coil A is made to rotate about a vertical axis (figure). No current flows in B if A is at rest. The current in coil A, when the current in B (at t-0) is counter-clockwise and the coil A is as shown at this instant, t=0, is

1. constant current clockwise

2. varying current clockwise

3. varying current counter clockwise

4. constant current counter clockwise

Q. 6. The self inductance L of a solenoid of length l and area of cross-section A, with a fixed number of turns N increases as

1. l and A increase

2. l decreases and A increases

3. l increases and A decreases

4. both l and A decrease

Q. 7. A metal plate is getting heated. It can be because

1. a direct current is passing through the plate

2. it is placed in a time varying magnetic field

3. it is placed in a space varying magnetic field, but does not vary with the time

4. a current (either direct or alternating) is passing through the plate

Q. 8. An emf is produced in a coil, which is not connected to an external voltage source. This can be due to

1. the coil being in a time varying magnetic field

2. the coil moving in a time varying magnetic field

3. the coil moving in a constant magnetic field

4. the coil is stationary in external spatially varying magnetic field, which does not change with time

Q. 9. The mutual inductance M_{12} of coil 1 with respect of coil 2

1. increases when they are brought nearer

2. depends on the current passing through the coils

3. increases when one of them is rotated about an axis

4. is the same as M_{21} of coil 2 with respect to coil 1

10. A circular coil expands radially in a region of magnetic field and no electromotive force is produced in the coil. This can be because

1. the magnetic field is constant

2. the magnetic field is in the same plane as the circular coil and it may or may not vary

3. the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably

4. there is a constant magnetic field in the perpendicular (to the plane of the coil) direction

Q. 3. A cylinder bar magnet is rotated about its axis. A wire is connected from the axis and is made to touch the cylindrical surface through contact. Then,

(a) a direct current flows in the ammeter A

(b) no current flows through the ammeter A\

(c) an alternating sinusoidal current flows through the ammeter A with a time $\mathrm{T}=\frac{2\mathrm{\pi}}{\mathrm{\omega}}$

(d) a time-varying non-sinusoidal current flows through the ammeter A

NEETprep Answer

Q. 4. There are two coils A and B as shown in the figure. A current starts flowing in B as shown when A is moved towards B and stops when A stops moving. The current in A is counterclockwise. B is kept stationary when A moves. We can infer that

(a) there is a constant current in the clockwise direction in A

(b) there is a varying current in A

(c) there is no current A

(d) there is a constant current in the counterclockwise direction in A

NEETprep Answer

Q. 5. Same as problem 4 except the coil A is made to rotate about a vertical axis (figure). No current flows in B if A is at rest. The current in coil A is as show at this instant, t=0, is

(a) constant current clockwise

(b) varying current clockwise

(c) varying current counter clockwise

(d) constant current counter clockwise

NEETprep Answer

Q. 6. The self-inductance L of a solenoid of length l and area of cross-section A, with a fixed number of turns N increases as

(a) l and A increase

(b) l decreases and A increases

(c) l increases and A decreases

(d) both l and A decrease

NEETprep Answer

Q. 7. A metal plate is getting heated. It can because

(a) a direct current is passing through the plate

(b) it is placed in a time-varying magnetic field

(c) it is placed in a space varying magnetic field, but does not vary with time

(d) a current (either direct or alternating) is passing through the plate

NEETprep Answer

Q. 8. An emf is produced in a coil, which is not connected to an external voltage source. This can be due to

(a) the coil is in a time-varying magnetic field

(b) the coil moving in a time-varying magnetic field

(c) the coil moving in a constant magnetic field

(d) the coil is stationary in an external spatially varying magnetic field, which does not change with time

NEETprep Answer

Q. 9. The mutual inductance ${\mathrm{M}}_{12}$ of coil 1 with respect to coil 2

(a) increases when they are bought nearer

(b) depends on the current passing through the coils

(c) increases when one of them is rotated about an axis

(d) is the same as ${\mathrm{M}}_{21}$ of 2 coil with respect to coil 1

NEETprep Answer

Q. 10. A circular coil expands rapidly in a region of the magnetic field and no electromotive force is produced in the coil. This can be because

(a) the magnetic field is constant

(b) the magnetic field is in the same plane as the circular coil and it may or may not vary

(c) the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably

(d) there is a constant magnetic field in the perpendicular (to the plane of the coil) direction

Q. 11. Consider a magnet surrounded by a wire with an on/off switch S(figure). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current flow in the circuit? Explain.

NEETprep AnswerQ. 12. A wire in the form of a tightly wound solenoid is connected to a DC source, and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain.

NEETprep AnswerQ. 13. A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain.

NEETprep AnswerQ. 14. Consider a metal ring kept on top of a fixed solenoid (say on cardboard) (figure). The centre of the ring coincides with the axis of the solenoid. If the current is suddenly switched on, the metal ring jumps up. Explain

NEETprep AnswerQ. 15. Consider a metal ring kept (supported by a cardboard) on top of a fixed solenoid carrying a current I. The centre of the ring coincides with the axis of the solenoid. If the current in the solenoid is switched off, what will happen to the ring?

NEETprep AnswerQ. 16. Consider a metallic pipe with an inner radius of 1 cm. If a cylindrical bar magnet of radius 0.8 cm is dropped through the pipe, it takes more time to come own than it takes for a similar unmagnetized cylindrical iron bar dropped through the metallic pipe. Explain.

NEETprep AnswerQ. 17. A magnetic field in a certain region is given by $\mathrm{B}={\mathrm{B}}_{0}\mathrm{cos}\left(\mathrm{\omega}\mathrm{t}\right)\hat{\mathrm{k}}$ and a coil of radius a with resistance R is placed in the x-y plane with its centre at the origin in the magnetic field (figure). Find the magnitude and the direction of the current at (a, 0, 0) at

NEETprep AnswerQ. 18. Consider a closed loop C in a magnetic field (figure). The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula $\mathrm{\varphi}={\mathrm{B}}_{1}{\mathrm{dA}}_{1},{\mathrm{B}}_{2}{\mathrm{dA}}_{2}$ ...... . Now, if we choose two different surfaces ${\mathrm{S}}_{1}\mathrm{and}{\mathrm{S}}_{2}$ having C as their edge, would we get the same answer for flux. Justify your answer.

NEETprep AnswerQ. 19. Find the current in the wire for the configuration shown in the figure. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper. $\mathrm{\theta}$ is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance d.

NEETprep AnswerQ. 20. A (current versus time) graph of the current passing through a solenoid is shown in figure. For which time is the back electromotive force (u) a maximum. If the back emf at t=3 s is e, find the back emf at t=7 s, 15s and 0 s. OA, AB and BC are straight line segments.

NEETprep AnswerQ. 21. There are two coils A and B separated by some distance. If a current f 2A flows through A, a magnetic flux of ${10}^{-2}$ Wb passes through B (no current through B). If no current passes through A and a current of 1 A passes through B, What is the flux through A?

Q. 22. A magnetic field $\mathrm{B}={\mathrm{B}}_{0}\mathrm{sin}\left(\mathrm{\omega t}\right)\hat{\mathrm{k}}$ covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d (figure). The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit? What is the force needed to keep the wire moving at constant velocity?

Q. 23. A conducting wire XY of mass m and negligible resistance slides smoothly on two parallel conducting wires as shown in figure. The closed-circuit has a resistance R due to AC. AB and CD are perfect conductors. There is a magnetic field B=B(t)$\hat{\mathrm{k}}$

(i) Write down the equation for the acceleration of the wire XY.

(ii) If B is independent of time, obtain v(t), assuming v (0)=${\mathrm{u}}_{0}$

(iii) For (ii), showing that the decrease in kinetic energy of XY equals the heat lost in.

Q. 24. ODBAC is a fixed rectangular conductor of negligible resistance (CO is not connected) and OP is a conductor which rotates clockwise with angular velocity $\mathrm{\omega}$ (figure). The entire system is in uniform magnetic field B whose direction is along the normal to the surface of the rectangular conductor ABDC. The conductor OP is in electric contact with ABDC. The rotating conductor has a resistance of $\mathrm{\lambda}$ per unit length. Find the current in the rotating conductor, as it rotates by $180\xb0$.

Q. 25. Consider an infinitely long wire carrying a current I(t), with $\frac{\mathrm{dI}}{\mathrm{dt}}=\lambda $=constant. Find the current produced in the rectangular loop of wire ABCD if its resistance is R (figure).

Q. 26. A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current $\mathrm{I}\left(\mathrm{t}\right)={\mathrm{I}}_{0}(1-\mathrm{t}/\mathrm{T})\mathrm{for}0\le \mathrm{t}\le \mathrm{T}$and I(0)=0 for t>T (figure). Find the total charge passing through a given point in the loop, in time T. THe resistance of the loop is R.

Q. 27. A magnetic field B is confined to a region r$\le $a and points out of the paper (the z-axis), r=0 being the centre of the circular region. Charged ring (charge=Q) of radius b, b>a and mass m lies in the x-y plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time $\u2206$t. Find the angular velocity $\mathrm{\omega}$ of the ring after field vanishes.

Q. 28. A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle $\mathrm{\theta}$ with respect to the horizontal (figure). The circuit is closed through a perfect conductor at the top. There is a constant magnetic field B along the vertical direction. If the rod is initially at rest, find the velocity of the rod as a function of time.

Q. 29. Find the current in the sliding rod AB (resistance=R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S i s closed at time t=0.

Q. 30. Find the current in the sliding rod AB (resistance=R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t=0.

Q. 31. A metallic ring of mass m and radius l (ring being horizontal) is falling under gravity in a region having a magnetic field. If z is the vertical direction, the z-component of the magnetic field is ${\mathrm{B}}_{\mathrm{z}}={\mathrm{B}}_{0}\left(1+\mathrm{\lambda}\mathrm{z}\right)$. If R is the resistance of the ring and if the ring has reached a constant velocity, use the conservation of energy to determine v in terms of m, B, $\mathrm{\lambda}$ and acceleration due to gravity g.

Q. 32. A long solenoid S and n turns per meter, with a diameter a. At the center of this coil, we place a smaller coil of N turns and diameter b (where b<a). If the current in the solenoid increases linearly, with time, what is the induced emf appearing in the smaller coil. Plot graph showing the nature of variation in emf, if current varies as a function of ${\mathrm{mt}}^{2}+\mathrm{C}$.